Matematické Fórum

stuart clark

If $I_{r} = \int^{e}_{1}\frac{1}{(1+\ln x)^r}dx.$ Then value of $\sum^{\infty}_{r=0}(1-rI_{r})$ is

check_drummer · 09. 05. 2019 20:35

↑ stuart clark:
Hi, why do you call the value $I_{r}$ when it doesn't depend on r?

stuart clark · 10. 05. 2019 04:06

Thanks ↑ check_drummer: i havr edited it.

Xellos · 31. 08. 2019 21:53

The argument of the sum seems to approach $-2/r$ , which means the sum is divergent. Not sure if that's what should happen, but unless I made a mistake in Python, it does happen.

jardofpr

hi ↑ Xellos:

Xellos napsal(a):
The argument of the sum seems to approach $-2/r$ , which means the sum is divergent

The sum argument approaching harmonic series argument does not necessarily imply divergence.

EDITED:
However it looks that in this case the series sum is little ahead of harmonics so
it seems that there only remains to show that it actually diverges analytically.